Evaluation of the integrals $ \int_{0}^{a}\frac{dx}{\sqrt{ax-x^2}} $ and $\int\frac{3x^5dx}{1+x^{12}}$

Question

I am struggling to find the values of these integrals after trying many substitution it did't worked for me

1) $$ \int_{0}^{a}\frac{dx}{\sqrt{ax-x^2}} $$ 2) $$ \int_{}^{}\frac{3x^5dx}{1+x^{12}} $$

Answer 1

HINT:

For $$\int_0^a\frac{dx}{\sqrt{ax-x^2}}=\int_0^a\frac{dx}{\sqrt{x(a-x)}}=\int_0^a\frac{dx}{|a|\sqrt{\frac xa(1-\frac xa)}}$$

Putting $x=a\sin^2y,dx=2a\sin y\cos y dy$ and as $x=0\implies y=0;x=a\implies y=\frac\pi2$

$$\int_0^a\frac{dx}{|a|\sqrt{\frac xa(1-\frac xa)}}=\int_0^{\frac\pi2}\frac{2a\sin y\cos y dy}{|a|\sin y\cos y}=\frac 2{\text{ sign }(a)}\int_0^{\frac\pi2}dy$$

$$\text{Similarly for the integral }\frac1{\sqrt{x(x-a)}},\text{ put } x=a\sec^2y$$

$$\text{Similarly for the integral }\frac1{\sqrt{x(x+a)}},\text{ put } x=a\tan^2y$$

Answer 2

Hint (a): Use change of variables, then trigonometric substitutions. \begin{align} \int_0^a \frac{dx}{\sqrt{ax-x^2}} = & \int_0^a \frac{dx}{\sqrt{\left(\frac{a}{2}\right)^2-\left(\frac{a}{2}\right)^2+2\cdot \left(\frac{a}{2}\right)x-x^2}} \\ = & \int_0^a \frac{\left(x-\left(\frac{a}{2}\right) \right)^\prime dx}{\sqrt{\left(\frac{a}{2}\right)^2-\left(x-\left(\frac{a}{2}\right) \right)^2}} \end{align} If $u=\left(x-\left(\frac{a}{2}\right) \right)$ then $du=\left(x-\left(\frac{a}{2}\right) \right)^\prime dx$ and we have the new integral \begin{align} \int_0^a \frac{dx}{\sqrt{ax-x^2}}=&\int_{\frac{a}{2}}^{\frac{3}{2}a} \frac{du}{\sqrt{\left(\frac{a}{2}\right)^2-u^2}} \end{align} Now you can proceed by trigonometric substitution $u=\left(\frac{a}{2}\right)\sin \theta$ then $ du=\left(\frac{a}{2}\right)\cos\theta \, d\theta $ and $$ \sqrt{\left(\frac{a}{2}\right)^2-u^2}= \sqrt{\left(\frac{a}{2}\right)^2-\left(\frac{a}{2}\right)^2\sin^2\theta}= \left|\frac{a}{2}\right|\cdot \left|\cos\theta\right| $$

Hint (b): Use change of variables and partial fractions in \begin{align} \int\frac{3x^5dx}{1+x^{12}} = & \int\frac{3}{6}\frac{(x^6)^\prime dx}{1+(x^{6})^2} \\ \end{align}

Answer 3

For the first integral, write

$$ax = \left (\frac{a}{2}\right)^2 - \left ( \frac{a}{2}-x\right)^2$$

and substitute $y = a/2-x$, then $y=(a/2) \sin{\theta}$. The result I get is $\pi$, independent of $a$.

For the second integral, substitute $y=x^6$, and use

$$\int \frac{dy}{1+y^2} = \arctan{y}+C$$